Problem: A curve in the plane is defined parametrically by the equations $x=\cos(\pi-t)$ and $y=\tan(2t)$. Find the value of $\dfrac{dy}{dx}$ at $t=\dfrac{\pi}{2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $-1$ (Choice C) C $-2$ (Choice D) D $2$
In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=\cos(\pi-t)$ and $y=\tan(2t)$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}\left(\tan(2t)\right)}{\dfrac{d}{dt}(\cos(\pi-t))} \\\\ &=\dfrac{2\sec^2(2t)}{\sin(\pi-t)} \end{aligned}$ Now let's evaluate $\dfrac{dy}{dx}$ at $t= {\dfrac{\pi}{2}}$ : $\begin{aligned} &\phantom{=}\dfrac{2\sec^2\left(2\left({\dfrac{\pi}{2}}\right)\right)}{\sin\left(\pi-{\dfrac{\pi}{2}}\right)} \\\\ &=\dfrac{2\sec^2(\pi)}{\sin\left(\dfrac{\pi}{2}\right)} \\\\ &=\dfrac{2\cdot 1}{1} \\\\ &=2 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $t=\dfrac{\pi}{2}$ is $2$.